C语言 unsigned long long 类型以十六进制格式打印
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unsigned long long type printing in hexadecimal format
提问by Paul the Pirate
I am trying to print out an unsigned long longlike this:
我正在尝试打印出unsigned long long这样的：
printf("Hex add is: 0x%ux ", hexAdd);
but I am getting type conversion errors since I have an unsigned long long.
但由于我有一个unsigned long long.
回答by gavinb
You can use the same llsize modifier for %x, thus:
您可以对 使用相同的ll大小修饰符%x，因此：
#include <stdio.h>
int main() {
unsigned long long x = 123456789012345ULL;
printf("%llx\n", x);
return 0;
}
The full range of conversion and formatting specifiers is in a great table here:
完整的转换和格式说明符在一个很棒的表格中：
回答by I?ya Bursov
try %llu- this will be long long unsigned in decimal form
尝试%llu- 这将是十进制形式的 long long unsigned
%llxprints long long unsigned in hex
%llx以十六进制打印 long long unsigned
回答by Joseph
printf("Hex add is: %llu", hexAdd);
回答by Cool Javelin
I had a similar issue with this using the MinGW libraries. I couldn't get it to recognize the %llu or %llx stuff.
我在使用 MinGW 库时遇到了类似的问题。我无法让它识别 %llu 或 %llx 的东西。
Here is my answer...
这是我的答案...
void PutValue64(uint64_t value) {
char a_string[25]; // 16 for the hex data, 2 for the 0x, 1 for the term, and some spare
uint32 MSB_part;
uint32 LSB_part;
MSB_part = value >> 32;
LSB_part= value & 0x00000000FFFFFFFF;
printf(a_string, "0x%04x%08x", MSB_part, LSB_part);
}
Note, I think the %04x can simply be %x, but the %08x is required.
请注意，我认为 %04x 可以简单地为 %x，但 %08x 是必需的。
Good luck. Mark
祝你好运。标记
